aramutselv

2021-12-27

Evaluate the indefinite integral.
$\int \frac{2{x}^{2}+x}{{\left(4{x}^{3}+3{x}^{2}\right)}^{2}}dx$

recoronarrv

Step 1
The given integral is,
$I=\int \frac{2{x}^{2}+x}{{\left(4{x}^{3}+3{x}^{2}\right)}^{2}}dx$
Using the substitution method, let $4{x}^{3}+3{x}^{2}=t$
on differentiating with respect to x on both side of the equation, we get
$\frac{d}{dx}\left(4{x}^{3}+3{x}^{2}\right)=\frac{dt}{dx}$
$4\left(3{x}^{2}\right)+3\left(2x\right)=\frac{dt}{dx}$
$12{x}^{2}+6x=\frac{dt}{dx}$
$6\left(2{x}^{2}+x\right)=\frac{dt}{dx}$
$2{x}^{2}+x=\frac{1}{6}\frac{dt}{dx}$
Therefore, the given integral becomes,
$I=\int \frac{1}{6{t}^{2}}dt$
Step 2
Using the power rule of integration, on solving the given integral, we get
$I=\int \frac{1}{6{t}^{2}}dt$
$=\frac{1}{6}\int {t}^{-2}dt$
$=\frac{1}{6}\left(\frac{{t}^{-2+1}}{-2+1}\right)+C$
$=-\frac{1}{6t}+C$
On putting back the value of t, we get
$I=-\frac{1}{6\left(4{x}^{3}+3{x}^{2}\right)}+C$
Therefore, the expression of the given integral is $-\frac{1}{6\left(4{x}^{3}+3{x}^{2}\right)}+C$

Chanell Sanborn

$\int \frac{2{x}^{2}+x}{{\left(4{x}^{3}+3{x}^{2}\right)}^{2}}dx$
$=\frac{1}{6}\int \frac{1}{{u}^{2}}du$
$\int \frac{1}{{u}^{2}}du$
$=-\frac{1}{u}$
$\frac{1}{6}\int \frac{1}{{u}^{2}}du$
$=-\frac{1}{6u}$
$=-\frac{1}{6\left(4{x}^{3}+3{x}^{2}\right)}$
$=-\frac{1}{6\left(4{x}^{3}+3{x}^{2}\right)}+C$
Lets

Vasquez